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\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\) => \(\frac{T}{2}=\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\)
=> \(T-\frac{T}{2}=\left(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\right)-\left(\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\right)\)
<=> \(\frac{T}{2}=\frac{2}{2^1}+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{2017}{2^{2016}}-\frac{2016}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
<=> \(\frac{T}{2}=1+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
Đặt: \(M=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}=>2M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\)
=> \(2M-M=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
=> \(M=\frac{1}{2}-\frac{1}{2^{2016}}< \frac{1}{2}\)
=> \(\frac{T}{2}< 1+\frac{1}{2}-\frac{2017}{2^{2017}}< 1+\frac{1}{2}=\frac{3}{2}\)
=> T < 3
Ta có :
\(S=\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{2016}{2017!}\)
\(S=\frac{3-1}{3!}+\frac{4-1}{4!}+\frac{5-1}{5!}+...+\frac{2017-1}{2017!}\)
\(S=\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+\frac{5}{5!}-\frac{1}{5!}+...+\frac{2017}{2017!}-\frac{1}{2017!}\)
\(S=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{4!}-\frac{1}{5!}+...+\frac{1}{2016!}-\frac{1}{2017!}\)
\(S=\frac{1}{2!}-\frac{1}{2017!}\)
\(S=\frac{1}{2}-\frac{1}{2017!}\)
Vậy \(S=\frac{1}{2}-\frac{1}{2017!}\)
Chúc bạn học tốt ~
s<2
bài này hình như mk lm ròi nhg ko nhớ là phải đáp án này ko
nếu sai cho mình xl
\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2017}\)
\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2017.2018}\)
\(\frac{1}{2}S=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(\frac{1}{2}S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(\frac{1}{2}S=\frac{1}{2}-\frac{1}{2018}\)
\(\frac{1}{2}S=\frac{504}{1009}\)
=> \(S=\frac{1008}{1009}\)
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